Selection Feb. 9, 2015 HUGEN 2022: Population Genetics J. Shaffer Dept. Human Genetics University of Pittsburgh

SelectionFeb. 9, 2015

HUGEN 2022: Population Genetics

J. ShafferDept. Human GeneticsUniversity of Pittsburgh

Objectives:

after this lecture you will need to be able to:

1. calculate and interpret fitness and selection coefficients 2. explain the qualitative (long-term) effects of different types

of selection (dominant, recessive, over-dominant, under-dominant) and of the combination of selection and mutation

3. calculate change in allele frequency 4. calculate over- and under-dominant equilibria5. interpret and make predictions about simulation plots

Hardy-Weinberg assumptions

• diploid organism• sexual reproduction• nonoverlapping generations• random mating• large population size• equal allele frequencies in the sexes• no migration• no mutation• no selection

The big picture: Evolution• Definition:

– change in the genetic composition (allele frequencies) of a population across successive generations

• Evolution vs. Hardy-Weinberg– the H-W Law tells us that if the assumptions are met, genotype and

allele frequencies do NOT change from one generation to the next– for evolution to occur, H-W assumptions must be violated– Which processes drive evolution?

• mutation

• natural selection• genetic drift

Selection - differing viability and/or fertility of different genotypes

Viability - some genotypes do not survive to birth or maturation

Fertility - some genotypes do not reproduce (as much)

This is an artificial division because ‘maturation’ can include fertility.

Definitions

wAA = P(AA individual reproduces)

wAa = P(Aa individual reproduces)

waa = P(aa individual reproduces)

Mathematical notation - fitness coefficients

wAA = P(AA individual reproduces)

wAa = P(Aa individual reproduces)

waa = P(aa individual reproduces)

We can’t really measure all three of these uniquely, so we usually set one of the w’s equal to 1, for example

wAA = 1

wAa = P(Aa reproduces) / P(AA reproduces)

waa = P(aa reproduces) / P(AA reproduces)

You can choose any of the three genotypes as the “reference.”

Mathematical notation - fitness coefficients

the selection coefficient, s, describes the degree of selection against the aa genotype

s = 1 – (waa / wAA)

the dominance coefficient, h, describes the degree of dominance.

h ͯ s = 1 – (wAa / wAA)

You can use either the w’s or h and s, and you should be able to translate back and forth freely.

Equivalently - selection coefficients

Recessive mutation (aa) that causes death in infancy

wAA = 1 (unaffected)

wAa = 1 (unaffected)

waa = 0 (die)

s = 1 – (waa / wAA) = 1 – (0 / 1) = 1

= 100% selection against aa

hs = 1 – (wAa / wAA) = 1 – (1 / 1) = 0 = 0% selection against Aa => h = 0

Example 1

Exam

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waa = 0 (die)

s = 1 – (waa / wAA) = 1 – (0 / 1) = 1



Example 1

Exam

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waa = 0 (die)

s = 1 – (waa / wAA) = 1 – (0 / 1) = 1



Example 1

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Aa mild disfigurement, extreme in aa- both have reduced rate of reproduction

wAA = 1 (unaffected)wAa = 0.9waa = 0.2

s = 1 – (waa / wAA) = 1 – (0.2 / 1) = 0.8 = 80% selection against aa

hs = 1 – (wAa / wAA) = 1 – (0.9 / 1) = 0.1 = 10% selection against Aa => h = 1/8

Example 2

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Example 2

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Example 2

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Huntington’s Disease - late onset! (Say “a” is the risk alllele.)

wAA = 1

wAa = 1

waa = 1

What if genetic testing?

Example 3

Exam

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Huntington’s Disease - late onset! (Say “a” is the risk alllele.)

wAA = 1

wAa = 1

waa = 1

What if genetic testing?

Example 3

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Sickle-cell anemia: aa: most die without treatmentw/ malaria Aa: survive malaria, unaffected by sickle-cell

AA: some do not survive malaria

wAA = 0.8 wAa = 1waa = 0.1

s = 1 – (waa / wAA) = 1 – (0.1 / 0.8) = 0.875 = 88% selection against aa

hs = 1 – (wAa / wAA) = 1 – (1 /0.8) = -0.25 = 25% selection in favor of Aa=> h = -0.286

What if there is no malaria?

Example 4: over-dominance

Exam

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wAA = 0.8 wAa = 1waa = 0.1





Exam

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wAA = 0.8 wAa = 1waa = 0.1





Exam

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wAA = 1 wAa = 1.25waa = 0.125


hs = 1 – (wAa / wAA) = 1 – (1.25 /1) = -0.25 = 25% selection in favor of Aa=> h = -0.286


Example 4: over-dominance II

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Females diploid: three genotype fitness coefficientsMales haploid two allele fitness coefficients

females maleswAA == 1 wA == 1wAa = 1 - hsf wa = 1 - sm

waa = 1 - sf

sf = 1 – (waa / wAA) sm = 1 – (wa / wA)

hsf = 1 – (wAa / wAA)

Example 5: X-linked

Special E

xampleassuming WAA and WA are

reference fitness coefficients

X-linked genotype freqpfpm, pfqm+pmqf, qfqm

pf, qf

Comments:any combinations of fitness scenarios are possible among females (i.e. recessive, over-dominant, etc.) which can occur in conjunction with two scenarios in males (equal/unequal).

Equilibria for X-linked fitness scenarios are outside the scope of this class; equilibria for diploid scenarios of over-/under-dominance and mutation-selection generally do not hold.

Clinical SelectionHuntington’s Disease dominant noneSickle cell (w. malaria) recessive over-dominantSickle cell (w/o. malaria) recessive recessiveHemochromatosis recessive noneAchondroplasia dominant intermediateCystic Fibrosis recessive intermediate

Clinical vs. Fitness disease types













Selection - major qualitative results

For dominant, recessive, or intermediate trait selection, the “risk” allele is eventually eliminated.

But, elimination is much quicker for dominant “risk” allele than recessive.

Recessive diseases are eliminated slowly because most “risk” alleles are in the heterozygotes, and there is no selection against them.

For over-dominance, an allele is not eliminated.

If we have ongoing new mutation, will an allele be eliminated?

Selection: quantitative stuff

p1 = # A alleles in the next generation

total # alleles in the next generation

denominator =average fitness ofthe population

think HWE freq. times fitness

p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

This formula is for the allele frequency, p1, of the “A” allele after exactly ONE generation. Can be recursively applied to calculate P(“A”) after more than one generation.


NOTE: these approximations assume that the average fitness is very close to 1, e.g., (p02)

(WAA) + (2p0q0)(WAa) + (q02)(Waa) ≈ 1. Therefore, they are valid if there is extreme selection

against rare genotypes, or light selection against common genotypes.

Approximations:

There is no simple formula for pt in terms of p0

ln(pt/qt) + (1/qt) ≈ ln(p0/q0) + 1/q0 + st 100% dominant

Allele frequency after t generations

ln(pt/qt) ≈ ln(p0/q0) + st/2 exactly additive

ln(pt/qt) - (1/pt) ≈ ln(p0/q0) - 1/p0 + st 100% recessive


What a genotype frequencies?

if random mating H-W assumption is held:

P(AA)t = pt2

P(Aa)t = 2ptqt P(aa)t = qt

2

as functions of allele frequencies, the genotype frequencies will change over time if allele frequencies change over time


X-linked genotype frequencies:

women: AA: pfpm Aa: pfqm+pmqf aa: qfqm

haplotypes in men:A: pf a: qf

pf1 = pf0pm0WAA + (1/2)(pf0qm0+pm0qf0)WAa

pf0pm0WAA + (pf0qm0+pm0qf0)WAa+ qf0qm0Waa

pm1 = pf0WA

pf0WA + qf0Wa

change after one generation:

Quantitative selection examplerecessive lethal disease, q0 = 0.04:

WAA = 1WAa = 1Waa = 0

Question: what are p1 and q1 in the next generation?

Exam

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p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

Exam

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p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

p1 = (0.962)(1) + (0.96)(0.04)(1)

(0.962)(1) + (2)(0.96)(0.04)(1) + (0.042)(0)

Exam

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p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

p1 = (0.962)(1) + (0.96)(0.04)(1)

(0.962)(1) + (2)(0.96)(0.04)(1) + (0.042)(0)

p1 = 0.9615

Exam

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q1 = 1 – 0.9615 =

1. very strong selection against aa, but allele frequency of a changes very little

2. most “risk” alleles are in heterozygotes3. frequency of A allele is high

p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

p1 = (0.962)(1) + (0.96)(0.04)(1)

(0.962)(1) + (2)(0.96)(0.04)(1) + (0.042)(0)

p1 = 0.9615

0.0385

Comments

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Selection: changes in allele frequencies

Two questions to explore:

1. how fast is an allele eliminated by selection?

2. what happens if there is over- or under-dominance?

How fast is an allele eliminated?

Change in allele frequencies over time depends on fitness or selection coefficients. In general…

- Stronger selection = faster elimination of risk allele- Dominant disease = faster elimination of risk allele- Recessive disease = slow elimination of risk allele(risk alleles “hiding” in heterozygotes)

How fast is an allele eliminated?

WAA = 1Waa = 0.95

generations

risk

alle

le f

requ

ency

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0 100 200 300 400 500

recessive

additive

dominant

Selection: changes in allele frequencies

Thwo questions to explore:

1. how fast is an allele eliminated by selection?

2. what happens if there is over- or under-dominance?

Overdominance

Overdominance: wAA < wAa > waa

If start “in the middle”, 0 < p < 1

peq(A) =

Overdominance equilibrium (under selection):

p = 1 then… stay at p = 1

stay at p = 0 If start at p = 0 then…

stable equilibrium

unstable equilibrium

wAa – waa

2wAa – wAA – waa

human example: sickle cell trait w/ malaria

Overdominance

p0(A) = 1

0 < p0(A) < 1

p0(A) = 0

stable equilibrium

Overdominance example

in the presence of warfarin

Warfarin: anti-coagulant used as rat-killer in WWII era

After many generations, the rats became resistant due to a mutation - resistant allele “R”; normal (wild-type) allele “S”

Given the following fitnesses of the different genotypes, what is the equilibrium frequency of “S”?

SS SR RRFitness 0.68 1.0 0.37

peq(A) =wAa – waa


peq(A) =1 – 0.37

2(1) – 0.68 – 0.37= 0.66

Underdominance

If start “in the middle”, 0 < p < 1

Underdominance: wAA > wAa < waa

Underdominance equilibrium:

p = 1 then… stay at p = 1

stay at p = 0 If start at p = 0 then…

unstable

If you start at the ‘middle’ equilibrium point, you stay there, otherwise you will go to p = 0 or p = 1

1.0 0.5 0.9

peq(A) =wAa – waa


stable

No good examples in humans• possibly rh factor

few examples in nature• lizard size• finches beak size

Underdominance

p0(A) = 1

0 < p0(A) < 1

p0(A) = 0

unstable equilibrium

Selection - things to know

calculation and interpretation of fitness and selection coefficients (w, s, h)

qualitative (long-term) effects of different types of selection (dominant, recessive, over-dominant, under-dominant) and of the combination of selection and mutation

calculation of change in allele frequency using:

calculation of over-dominant and under-dominant equilibria

p1 = (p0

2)(WAA) + (1p0q0)(WAa)

(p02)(WAA) + (2p0q0)(WAa) + (q0

2)(Waa)

peq(A) =wAa – waa


Documents

Selection Feb. 9, 2015 HUGEN 2022: Population Genetics J. Shaffer Dept. Human Genetics University of Pittsburgh